Trig graphs, reciprocal and inverse trig functions, basic identities – A-level Maths exam questions with mark scheme

1StandardCalculator5 marks

Solve, for $0\leqslant\theta<360^\circ$, the equation $$2\tan^{2}\theta+\sec\theta=1$$ giving your answers to one decimal place where appropriate. [5]

Uses $\tan^{2}\theta=\sec^{2}\theta-1$M1
$2\sec^{2}\theta+\sec\theta-3=0$A1
Solves the 3TQ, $(2\sec\theta+3)(\sec\theta-1)=0$, and uses $\cos\theta=\dfrac{1}{\sec\theta}$ to obtain $\cos\theta=1$ or $\cos\theta=-\tfrac23$M1
Two correct answers from $0^\circ,\ 131.8^\circ,\ 228.2^\circ$A1
All three: $0^\circ,\ 131.8^\circ,\ 228.2^\circ$ and no extras in rangeA1
Answer: $\theta=0^\circ,\ 131.8^\circ,\ 228.2^\circ$
Examiner tip: $\theta=0^\circ$ is in the range $0\leqslant\theta<360^\circ$ – it is the answer most often missed.

2HardCalculator6 marks

(a) Show that the equation $$4\sin x\tan x=15$$ can be written in the form $$4\cos^{2}x+15\cos x-4=0.$$ [3]

(b) Hence solve, for $0\leqslant x<360^\circ$, $$4\sin x\tan x=15,$$ giving your answers to 1 decimal place. [3]

(a) for using $\tan x=\dfrac{\sin x}{\cos x}$ to obtain $\dfrac{4\sin^{2}x}{\cos x}=15$M1
(a) for multiplying by $\cos x$ and using $\sin^{2}x=1-\cos^{2}x$M1
(a) $4-4\cos^{2}x=15\cos x\Rightarrow4\cos^{2}x+15\cos x-4=0$ with no errorsA1*
(b) for solving the quadratic: $(4\cos x-1)(\cos x+4)=0$, giving $\cos x=\tfrac14$M1
(b) $75.5^\circ$ (awrt)A1
(b) $284.5^\circ$ (awrt), with $\cos x=-4$ rejected or ignored and no extra solutionsA1
Answer: (b) $x=75.5^\circ,\ 284.5^\circ$
Examiner tip: State that $\cos x=-4$ has no solutions (since $-1\leqslant\cos x\leqslant1$) rather than leaving it hanging.

3StretchCalculator8 marks

The depth of water, $d$ metres, at a harbour entrance is modelled by $$d=4+1.5\sin(30t)^\circ,\qquad 0\leqslant t\leqslant24,$$ where $t$ is the time in hours after midnight. A sketch of the graph is shown.

(a) State the maximum depth of water predicted by the model. [1]

(b) Find the value of $t$ when this maximum first occurs. [1]

A boat can only pass the entrance when the depth is more than $4.75$ metres.

(d) Find the total length of time between $t=12$ and $t=24$ during which the boat can pass. [2]

(a) $5.5$ mB1
(b) $t=3$B1
(c) for $4+1.5\sin(30t)=4.75\Rightarrow\sin(30t)=0.5$M1
(c) $30t=30^\circ$ and $150^\circ$A1
(c) for dividing their angles by 30dM1
(c) $t=1$ and $t=5$ (1 am and 5 am)A1
(d) for using periodicity (period 12 hours) to find $t=13$ and $t=17$, or $30t=390^\circ,510^\circ$M1
(d) $4$ hoursA1
Answer: (a) $5.5$ m (b) $t=3$ (c) $t=1,\ 5$ (d) $4$ hours
Examiner tip: Multiply out the interval: $0\leqslant t\leqslant12$ means $0\leqslant30t\leqslant360^\circ$, so look for every angle with $\sin=0.5$ up to $360^\circ$.

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